Problem: Nastya likes reading and even spends whole days in a library sometimes. Today she found a chronicle of Byteland in the library, and it stated that there lived shamans long time ago. It is known that at every moment there was exactly one shaman in Byteland, and there were n shamans in total enumerated with integers from 1 to n in the order they lived. Also, each shaman had a magic power which can now be expressed as an integer.

The chronicle includes a list of powers of the n shamans. Also, some shamans can be king-shamans, if they gathered all the power of their predecessors, i.e. their power is exactly the sum of powers of all previous shamans. Nastya is interested in whether there was at least one king-shaman in Byteland.

Unfortunately many of the powers are unreadable in the list, so Nastya is doing the following:

  * Initially she supposes some power for each shaman. 
  * After that she changes the power of some shaman q times (the shamans can differ) and after that wants to check if there is at least one king-shaman in the list. If yes, she wants to know the index of any king-shaman. 



Unfortunately the list is too large and Nastya wants you to help her.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2·105).

The second line contains n integers a1, ..., an (0 ≤ ai ≤ 109), where ai is the magic power of the i-th shaman.

After that q lines follow, the i-th of them contains two integers pi and xi (1 ≤ pi ≤ n, 0 ≤ xi ≤ 109) that mean that the new power of the pi-th shaman is xi.

Output

Print q lines, the i-th of them should contain  - 1, if after the i-th change there are no shaman-kings, and otherwise a single integer j, where j is an index of some king-shaman after the i-th change.

If there are multiple king-shamans after each change, print the index of any of them.

Examples

Input

2 1
1 3
1 2


Output

-1


Input

3 4
2 2 3
1 1
1 2
2 4
3 6


Output

3
2
-1
3


Input

10 7
0 3 1 4 6 2 7 8 10 1
2 5
1 3
9 36
4 10
4 9
1 2
1 0


Output

1
-1
9
-1
4
-1
1

Note

In the first example powers of shamans after the first change are equal to (2, 3). The answer equals  - 1, because the sum of powers of shamans before the first shaman is equal to 0, and before the second is equal to 2.

In the second example after the first change the powers are equal to (1, 2, 3). The answer is equal to 3, because the power of the third shaman is equal to 3, and the sum of powers of the first and the second shaman is also 1 + 2 = 3. After the second change the powers become equal to (2, 2, 3), where the answer equals 2. After the third change the powers become equal to (2, 4, 3), where the answer equals  - 1. After the fourth change the powers become equal to (2, 4, 6), where the answer equals 3.

Solution (in PYTHON):

Solution:
class maxseg:
    def __init__(self,data):
        n = len(data)
        m = 1
        while m<n:m*=2
        self.n = n
        self.m = m

        self.data = [0]*(m+m)

        for i in range(n):
            self.data[m+i]=data[i]
        for i in reversed(range(m)):
            self.data[i]=max(self.data[2*i],self.data[2*i+1])

    def setter(self,i,val):
        i += self.m
        self.data[i]=val
        i//=2
        while i>0:
            a = max(self.data[2*i],self.data[2*i+1])
            if self.data[i]==a:break
            self.data[i]=a
            i//=2

    # First time A[i]>=val
    def first_time(self,l,r,val):
        l+=self.m
        r+=self.m

        segs1 = []
        segs2 = []
        while l<r:
            if l%2==1:
                segs1.append(l)
                l+=1
            if r%2==1:
                r-=1
                segs2.append(r)
            l//=2
            r//=2
        segs = segs1+segs2[::-1]
        goal = None
        for seg in segs:
            if self.data[seg]>=val:
                while seg<self.m:
                    if self.data[2*seg]>=val:
                        seg *= 2
                    else:
                        seg = 2*seg+1
                goal = seg-self.m
                break
        return goal

class sumseg:
    def __init__(self,data):
        n = len(data)
        m = 1
        while m<n:m*=2
        self.n = n
        self.m = m
        self.data = [0]*(m+m)
        for i in range(n):
            self.data[m+i]=data[i]
        for i in reversed(range(m)):
            self.data[i]=self.data[2*i]+self.data[2*i+1]

    def setter(self,i,val):
        i += self.m
        diff = val-self.data[i]
        while i>0:
            self.data[i]+=diff
            i//=2
    def summa(self,l,r):
        l+=self.m
        r+=self.m
        s = 0
        while l<r:
            if l%2==1:
                s += self.data[l]
                l+=1
            if r%2==1:
                r-=1
                s += self.data[r]
            l//=2
            r//=2
        return s

def main():
    n,q = [int(x) for x in input().split()]
    
    A = [int(x) for x in input().split()]
    
    maxA = maxseg(A)
    sumA = sumseg(A)
    
    out = []
    for _ in range(q):
        p,x = [int(x) for x in input().split()]
        p-=1
    
        A[p]=x
        maxA.setter(p,x)
        sumA.setter(p,x)
    
        i=0
        lista = []
        while i!=None:
            s = sumA.summa(0,i)
            if A[i]==s:
                lista.append(i)
                break
            s += A[i]
            i = maxA.first_time(i+1,n,s)
        if len(lista)==0:
            out.append('-1')
        else:
            out.append(' '.join(str(x+1) for x in lista))
    print '\n'.join(out)
######## Python 2 and 3 footer by Pajenegod and c1729

# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.

# So on cf, use PyPy2 for best string performance.

py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange

import os, sys
from io import IOBase, BytesIO

BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None

    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s

    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()

    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            self.write = lambda s:self.buffer.write(s.encode('ascii'))
            self.read = lambda:self.buffer.read().decode('ascii')
            self.readline = lambda:self.buffer.readline().decode('ascii')


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n')

# Cout implemented in Python
import sys
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'

# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'0' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'0' [0]:
        A.append(sign*numb)
    return A

if __name__== "__main__":
    main()